What is the foci (x^2)/(16)+(y^2)/(49)=1 ?

The foci (x^2)/(16)+(y^2)/(49)=1 is (0,sqrt(33)),(0,-sqrt(33))

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foci (x^2)/(16)+(y^2)/(49)=1

foci

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

(0, 33), (0, - 33)

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 4

(0, 0 + c), (0, 0 - c)

Compute

c :

c = 7^{2} - 4^{2} : 33

(0, 0 + 33), (0, 0 - 33)

Refine

(0, 33), (0, - 33)

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

7 x^{2} + 2 y^{2} = 14

25 x^{2} + 36 y^{2} = 900

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

What is the foci (x^2)/(16)+(y^2)/(49)=1 ?
The foci (x^2)/(16)+(y^2)/(49)=1 is (0,sqrt(33)),(0,-sqrt(33))